Question

One of the equal sides of an isosceles triangle is 13cm and its perimeter is 50cm. Find the area of the triangle

Answer 1

In Isosceles $\Delta ABC$

$AB=AC=13cm$ But perimeter $=50cm$

$BC=50-(13+13cm)$

$=50-26=24cm$

$AD\perp BC$

$BD=DC=\frac{24}{2}=12cm$

In right $\Delta ABD$

$A{B}^2=A{D}^2+B{D}^2\left(PythagorasTheorem\right)$

$(13{)}^2=A{D}^2+(12{)}^2$

$\Rightarrow 169=A{D}^2+144$

$\Rightarrow A{D}^2=169-144=25=(5{)}^2$

$AD=5cm$

Area of $\Delta ABD$ $=\frac{1}{2}Base\times Altitue=\frac{1}{2}\times BC\times AD$

$\frac{1}{2}\times 24\times 5=60c{m}^2$